Which statements correctly describe the formula or name of a compound? Select all that apply.
OA. The formula of nitrogen trifluoride is NF 3
B. The formula of ammonia is NH3.
C. The name of AlF, is trialuminum fluoride.
D. The formula of calcium chloride is CaCl2
E. The name of Li, Se is lithium selenate.
OF. The formula of dinitrogen monoxide is NO
2
Results
G. The formula of sulfur trioxide is 30.
Cho
Аа
H. The formula of magnesium hydroxide is Mg(OH)2
G.

Answer:

a) The ideal speed = 16.21 m/s

b) Minimum co-efficient of friction = 0.216

Explanation:

From the given information:

The ideal speed can be determined by considering the centrifugal force component and the gravity component.

[tex]\dfrac{mv^2}{r}cos \theta = mg sin \theta[/tex]

[tex]v = \sqrt {gr \ tan \theta}[/tex]

[tex]= \sqrt{(9.8 \ m/s^2) (100) \ tan 15^0}[/tex]

= 16.21 m/s

(b)

Let assume that it requires 25 km/h to take the same curve.

Then, using the equilibrium conditions;

[tex]mg \ sin \theta = \dfrac{mv^2}{r} cos \theta + \mu ((\dfrac{mv^2}{r}) sin \theta + mg cos \theta)[/tex]

[tex]\mu = \dfrac{mg sin \theta - \dfrac{mv^2}{r} cos \theta }{((\dfrac{mv^2}{r}) sin \theta + mg cos \theta) }[/tex]

[tex]\mu = \dfrac{g sin \theta - \dfrac{ v^2}{r} cos \theta }{((\dfrac{v^2}{r}) sin \theta + g cos \theta) }[/tex]

[tex]\mu = \dfrac{(9.8 \ m/s^2 ) sin (15^0) - \dfrac{ \dfrac{(25 \times 10^3}{3600} \ m/s)^2 }{100 \ m } cos (15^0) }{((\dfrac{(\dfrac{25 \times 10^3}{3600} )^2}{100}) sin 15^0 + (9.8 \ m/s^2) cos 15^0 ) }[/tex]

[tex]\mathbf{\mu = 0.216}[/tex]

Answer:

4

Units:

Newtons

Answer: the total energy of the puck, ice surface, and surrounding air decreases to zero

Explanation:

A hockey puck slides across the ice and eventually comes to a stop because of friction between surface of the puck and ice surface.

Friction is a resistance to motion of the object. for example, when a body slides on horizontal surface in positive x direction, it has friction in negative x direction and that measure of friction is a frictional force. frictional force is directly proportional to the Normal(N). i.e. [tex]F_{fri}[/tex] ∝ N

[tex]F_{fri}[/tex] = μN where μ is called as coefficient of the friction. It is a dimensionless quantity.

When a body is kept on horizontal surface, its normal will be straight upward which is reaction of mg. i.e. N=mg.

Frictional force is equal to

[tex]F_{fri}[/tex] = μmg

When hockey puck slides across the ice, friction between surface of puck and surface of ice produces resistance to the motion of the puck, due to resistance puck slow down slowly and eventually come to a stop. Motion and frictional force are opposite to each other. we can calculated the exact value of frictional force when we know the coefficient of friction between puck and ice surface.

hence due to friction, puck come to a stop

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Answer:

I. The change in temperature of the object.

II. The initial volume of the object.

Explanation:

Generally, the volume of a solid object tends to increase as its temperature increases and this phenomenon is known as thermal expansion.

Hence, the quantities which determine how large the change in volume of a solid object is includes;

I. The change in temperature of the object.

II. The initial volume of the object.

This ultimately implies that, when a solid object is heated, the atoms of the object vibrate rapidly about their fixed points and thus, causing an increase in the volume of the object.

In conclusion, this scientific phenomenon known as thermal expansion is valid and true for all the three (3) states of matter;

Answer:

51.04 joules

Explanation:

The movement by the gloves = 11cm

Force by the baseball = 464N

First of all we are going to convert cm to metres

11cm = 11/100 meters = 0.11m

The formula for workdone is given as:

W = f x d

W = workdone

F = force

D = distance

Workdone = 464 x 0.11

= 51.04 joules

The workdone by the ball is 51.04 joules

Thank you

Answer:

-22.1

Explanation:

1 / 4

Step 1. List horizontal (xxx) and vertical (yyy) variables

xxx-direction yyy-direction

t=\text?t=?t, equals, start text, question mark, end text t=\text?t=?t, equals, start text, question mark, end text

a_x=0a  

x

​  

=0a, start subscript, x, end subscript, equals, 0 a_y=-9.8\,\dfrac{\text m}{\text s^2}a  

y

​  

=−9.8  

s  

2

m

​  

a, start subscript, y, end subscript, equals, minus, 9, point, 8, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction

\Delta x=255\,\text mΔx=255mdelta, x, equals, 255, start text, m, end text \Delta y=\text ?Δy=?delta, y, equals, start text, question mark, end text

v_x=v_{0x}v  

x

​  

=v  

0x

​  

v, start subscript, x, end subscript, equals, v, start subscript, 0, x, end subscript v_y=?v  

y

​  

=?v, start subscript, y, end subscript, equals, question mark

v_{0x}=120\,\dfrac{\text m}{\text s}v  

0x

​  

=120  

s

m

​  

v, start subscript, 0, x, end subscript, equals, 120, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction v_{0y}=0v  

0y

​  

=0v, start subscript, 0, y, end subscript, equals, 0

Note that there is no horizontal acceleration, so v_x=v_{0x}v  

x

​  

=v  

0x

​  

v, start subscript, x, end subscript, equals, v, start subscript, 0, x, end subscript. The time is the same for the xxx and yyy directions.

Also, the package has no initial vertical velocity.

Our yyy-direction variable list has too many unknowns to solve for \Delta yΔydelta, y directly. Since both the yyy- and xxx-directions have the same time ttt and horizontal acceleration is zero, we can solve for ttt from the xxx-direction motion by using equation:

\Delta x=v_xtΔx=v  

x

​  

tdelta, x, equals, v, start subscript, x, end subscript, t

Once we know ttt, we can solve for \Delta yΔydelta, y using the kinematic equation that does not include the unknown variable v_yv  

y

​  

v, start subscript, y, end subscript:

\Delta y=v_{0y}t+\dfrac {1}{2}a_yt^2Δy=v  

0y

​  

t+  

2

1

​  

a  

y

​  

t  

2

delta, y, equals, v, start subscript, 0, y, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, start subscript, y, end subscript, t, squared

Hint #22 / 4

Step 2. Find ttt from horizontal variables

\begin{aligned}\Delta x&=v_xt \\\\ t&=\dfrac{\Delta x}{v_{0x}} \\\\ t&=\dfrac{255\,\text m}{120\dfrac{\text m}{\text s}} \\\\ &=2.125\,\text s \end{aligned}  

Δx

t

t

​  

=v  

x

​  

t

=  

v  

0x

​  

Δx

​  

=  

120  

s

m

​  

255m

​  

=2.125s

​  

Hint #33 / 4

Step 3. Find \Delta yΔydelta, y using ttt

Using ttt to solve for \Delta yΔydelta, y gives:

\begin{aligned}\Delta y&=v_{0y}t+\dfrac{1}{2}a_yt^2 \\\\ &=\cancel{ (0 )t}+\dfrac{1}{2}\left (-9.8\dfrac{\text m}{\text s^2}\right )\left(2.125\,\text s\right)^2 \\\\ &=-22.1\,\text m \end{aligned}  

Δy

​  

=v  

0y

​  

t+  

2

1

​  

a  

y

​  

t  

2

=  

(0)t

​  

+  

2

1

​  

(−9.8  

s  

2

m

​  

)(2.125s)  

2

=−22.1m

​  

Hint #44 / 4

The correct answer is -22.1\,\text m−22.1mminus, 22, point, 1, start text, m, end text.

Answer:

Aluminum

Explanation:

promise

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Answer:aluminum

Explanation:

Answer: The bones won't fracture.

Explanation: Stress, in Physics, is a quantity describing forces that can cause deformation. Strain is the measure of how muc an object can be stretched or deformed. The ratio between stress and strain is called Young's modulus or elastic modulus

Breaking Stress of Bone is the maximum stress a bone can take before a rupture occur.

To determine if a person will break his/her bones by jumping from a height, we determine the energy necessary for that jump and compare it with the energy necessary to break a bone.

The energy for breaking a bone is calculated as

[tex]E=\frac{Al_{0}\sigma_{B}^{2}}{2Y}[/tex]

A is the area in m²

l₀ is length in m

[tex]\sigma_{B}[/tex] is breaking stress in N/m²

Y is Young's modulus in N/m²

Calculating energy to break a bone:

[tex]E=\frac{4.10^{-4}.7.10^{-1}.(1.5.10^{8})^{2}}{2.(1.5.10^{10})}[/tex]

[tex]E=210[/tex] J

This is the energy necessary to break one leg bone, so as there are 2, energy will be 420 Joules.

Potential energy gained by jumping is calculated as

E = m.g.h

m is mass in kg

g is acceleration due to gravity in m/s²

h is height in m

Calculating

E = 80.(9.8)(0.3)

E = 235.2 J

Comparing the two energies, potential energy for jumping is less than maximum energy a bone can absorve without breaking, so the leg bones won't suffer a fracture.

Answer:

the answer is b luv .

Explanation:

Doppler ultrasound would be most useful in detecting a blockage in a heart artery.

By monitoring the rate of change in pitch, a Doppler ultrasound may calculate how quickly blood flows (frequency). A sonographer with training in ultrasound imaging applies pressure to your skin with a tiny, hand-held instrument (transducer) roughly the size of a bar of soap across the area of your body being scanned, moving from one place to another as required.

As an alternative to more invasive treatments like angiography, which involves injecting dye into the blood arteries to make them visible on X-ray images, this test may be performed.

Your doctor may use a Doppler ultrasound to assess for artery damage or to keep track of specific vein and artery therapies.

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Answer:

6kg

Explanation:

According to conservation of kinetic energy

m1u1+m2u2 = (m1+m2)v

m1 and m2 are the masses of the bodies

u1 and u2 are the initial velocities

v is the final velocity

Given

m1 =2kg

u1 = 8.0m/s

m2 = ?

u2 = 0m/s (second ice at rest)

v = 2.0m/s

Substitute into the formula

2(8)+m2(0) = (2+m2)(2)

16+0 = 4+2m2

16-4= 2m2

12 = 2m2

m2 =12/2

m2 = 6kg

Hence the mass of the second block is 6kg

Answer:

1.25 m

Explanation:

This is the vertical height not the distance along the slope.

[tex]K=U\\\frac{1}{2}mv^{2} = mgh\\h = \frac{v^{2}}{2g}=\frac{25}{20}=1.25 m[/tex]

The height the car can reach if the the track starts to climb upwards is 1.2742 meters up.

Kinetic energy is energy possessed by a body by virtue of its movement. Potential energy is the energy possessed by a body by virtue of its position or its relation with its surrounding systems.

P.E. = mass × g × height

K.E. = 0.5 × mass × (velocity)²

Given that the toy car has a mass of 0.025 kg and is traveling on a horizontal track with a velocity of 5 m/s. Now, the car starts to climb up vertically, therefore, the kinetic energy will be converted to potential energy.

Kinetic Energy = Potential Energy

0.5 × mass × (velocity)² = mass × g × height

Cancel mass from both the sides,

0.5 × (velocity)² = g × height

0.5 × (5 m/s)² = 9.81 m/sec² × height

height = 1.2742 meters

Hence, the car will travel 1.2742 meters up.

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Answer:let initial velocity u=14m/s

Final velocity v=20m/s

Time taken t=30

Acceleration =a

V=u +at

a= (20-14)/30

a=0.2m/s^2

Explanation:

Acceleration is the change in velocity with respect to time.

The magnitude of the vertical component of the projectile's initial velocity is 200 m/s × sin 30°.

The diagrammatic representation of the velocity of the projectile can be seen in the attached image below.

From the diagram, let consider the ΔOAP where Vector OP makes an ∠θ = 30° to the horizontal x-axis.

where;

∴

Using trigonometric approach for ΔOAP;

[tex]\mathbf{sin\theta = \dfrac{AP}{OP}}[/tex]

[tex]\mathbf{AP =OP\times sin \theta}}[/tex]

AP = 200 × sin 30°

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Answer:

b) stopping the object

Explanation:

Friction always slows a moving object down. ... Friction can be a useful force because it prevents our shoes slipping on the pavement when we walk and stops car tyres skidding on the road. When you walk, friction is caused between the tread on shoes and the ground. This friction acts to grip the ground and prevent sliding

Answer:

The BTU, or British thermal unit, is actually a measure of heat.